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2 years ago

ERROR ANALYSIS Describe and correct the error in

Mathematics

1 answer:

valkas [14]2 years ago

3 0

Answer:

The graph should be stretched rather than become narrower.

Step-by-step explanation:

To figure this out, just create some example points.

At x = 0, your y-value will always be 0. But if you were to plug in the value 1, you would get a y-value of 1 in y=x^2, but a value of 0.5 in y=0.5x^2. If you were to plug in a value of 2, you would get a value of 4 in y=x^2, but a value of 2 in y=0.5x^2.

If you continue this pattern for a few more points, then plot them, you will see that adding a coefficient of 0.5 simply stretches the graph

You might be interested in

Multiply 4/5 by the reciprocal of −2/10.

Svetach [21]

Answer:

4/5*reciprocal of -2/10

The reciprocal of -2/10 is 10/-2

4/5*10/2

Cancelling the numbers

-4 is the answer

Step-by-step explanation:

I hope it will help you :)

7 0

3 years ago

Read 2 more answers

Select the functions that have identical graphs.

Elan Coil [88]

Answer:

c. 1 and 3

Step-by-step explanation:

To quickly solve this problem, we can use a graphing tool or a calculator to plot each equation.

Please see the attached image below, to find more information about the graph

The equations are:

1) y = sin (3x + π/6)

2) y = cos (3x - π/6)

3) y = cos (3x - π/3)

Looking at the graphs, we can see that the identical ones

are equations one and three

Correct option:

c. 1 and 3

3 0

3 years ago

Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i

OLEGan [10]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

$H_0:\mu=18\\\\H_a:\mu\neq18$

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size : n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : $\overline{x}=18.8$

Standard deviation : $\sigma=2.8$

Test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02$

The p-value= $2P(z>2.02)=0.0433834$

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0

3 years ago

There were 436 tickets purchased for a major league baseball game. The general admission tickets cost $6.50 and the upper reser

monitta

Answer:

136 general admission tickets were purchased, and 300 upper reserved tickets were purchased.

Step-by-step explanation:

This question is solved using a system of equations.

I am going to say that:

x is the number of general admission tickets purchased.

y is the number of reserved tickets purchased.

There were 436 tickets purchased for a major league baseball game.